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Each test is based on a null hypothesis, which is that there is no difference between the

means of the two data sets. The tests measure how likely the hypothesis is to be true.

The attraction of such tests is that they are easy to carry out and interpret.

Analysis of a standard solution – one-samplet-test

Once the choice of the analytical method to be used for a particular biochemical assay

has been made, the normal first step is to carry out an evaluation of the method in the

laboratory. This evaluation entails the replicated analysis of a known standard

solutionof the test analyte and the calculation of the mean and standard deviation

of the resulting data set. The question is then asked ‘Does the mean of the analytical

results agree with the known value of the standard solution within experimental

error?’ To answer this question at-testis applied.

In the case of the analysis of a standard solution the calculated mean and standard

deviation of the analytical results are used to calculate a value of the Student’s

t(tcalc) using equation 1.15. It is then compared with table values oft(ttable) for the

particular degrees of freedom of the data set and at the required confidence level

(Table 1.7).

tcalc¼ðknown valuexÞ

s

pn

ð 1 : 15 Þ

These table values oftrepresent critical values that separate the border between

different probability levels. Iftcalcis greater thanttablethe analytical results are

deemed not to be from the same data set as the known standard solution at the

selected confidence level. In such cases the conclusion is therefore drawn that the

analytical results do not agree with the standard solution and hence that there are

unidentified errors in them. There would be no point in applying the analytical

method to unknown test analyte samples until the problem has been resolved.

Example 6IDENTIFICATION OF AN OUTLIER EXPERIMENTAL RESULT

Question If the data set in Example 5 contained an additional value of 3.0 mM, could this
value be regarded as an outlier point at the 95% confidence level?

Answer From equation 1.15

Qexp¼

3 : 0  2 : 6

3 : 0  2 : 2

¼

0 : 4

0 : 8

¼ 0 : 5

Using Table 1.11 for six data pointsQtableis equal to 0.62.

SinceQexpis smaller thanQtablethe point should not be rejected as there is

more than a 95% chance that it is part of the same data set as the other five values.

It is easy to show that an additional data point of 3.3 rather than 3.0 mM would

give aQexpof 0.64 and could be rejected.

28 Basic principles